Suppose that $m(\lim_{x\to \infty}f_n)=f$ in $[a,b]$. i.e $(f_n)$ is a sequence of integrable functions and $f$ is a Riemann integrable function, such that $(f_n)$ converges in measure to $f$ in [a,b]. If $g$ is Riemann Integrable in $[a,b]$ and we define
$$h(x) = \int_a^xf(t)g(t)dt,$$ $$ h_n(x) = \int_a^xf_n(t)g(t)dt$$ for $x \in [a,b].$ Prove that $h_n \to h$ uniformly in $[a,b]$.
I am lost with this problem, and would like to make the homework. Today I read the definition of convergence in measure, and some info that concerns it. Some related theorems, for example.I know the definition of something that converges uniformly to other something, but I have no idea what to do, how to think about this problem, or even what kind of example could serve for me to see what I want to prove, or how would I go about proving the statement.
Any help with this would be very appreciated.
Note that $f_n\to f(\text{in measure $\mu$})\implies $
$\lim_{n\to \infty}\mu (\{x\in [a,b]:|f_n(x)-f(x)|\ge \epsilon\})=0 \forall \epsilon>0$
Since $g$ is Riemann-Integrable $\implies |g|\le M$
Now $|\int _a^x f_n(t)g(t)-f(t)g(t)|\le \int _a^x|f_n(t)-f(t)||g(t)|\le M \int _a^x|f_n(t)-f(t)|$
Now consider for any $\epsilon >0$;
$A=\{x\in [a,b]:|f_n(x)-f(x)|< \epsilon\}$ and $B=\{x\in [a,b]:|f_n(x)-f(x)|\ge \epsilon\}$
$\int _a^x|f_n(t)-f(t)|=\int _{A\cup B} |f_n(t)-f(t)|$
Note that $\mu(B)=0 $ and hence $\int _a^x|f_n(t)-f(t)|<(b-a)\epsilon$
Hence $|\int _a^x f_n(t)g(t)-f(t)g(t)|<M(b-a)\epsilon$ (converging uniformly)